Just like the classic 4 prisoners hats riddle, here we have 6 prisoners buried to their necks in the ground. They can only look straight ahead so that A only sees B, C, D, E while B sees C, D, E, and so on and F is completely hidden from view. The warden gives them each hats and tells them that there are 3 red hats and 3 white hats. The warden also tells them that he has cut out one prisoner's tongue (in this case C) so that he cannot speak at all (the mute knows that he is mute). All prisoners are executed if they make any noise other than to clearly announce their own hat color. If a prisoner answers correctly, all prisoners will be set free. If incorrectly they will all be executed.


One prisoner will be able to say his own hat color with certainty. Which one?

To clear up some confusion:
1) No prisoner knows who the mute is except the mute himself.
2) The picture is how the story actually went down.
3) The lateral thinking tag was added just because the solution takes some time-dimensional thinking.

  • $\begingroup$ If they all guess red or white they will be set free. $\endgroup$ – Yout Ried Sep 5 '18 at 2:50
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    $\begingroup$ They had no time beforehand to devise that strategy. $\endgroup$ – tyobrien Sep 5 '18 at 2:53
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    $\begingroup$ Does anyone other than C know that C is mute? $\endgroup$ – LeppyR64 Sep 5 '18 at 4:08
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    $\begingroup$ Is your image an example or the actual way the hats are distributed? $\endgroup$ – Mark Sep 5 '18 at 8:33
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    $\begingroup$ @DanielP The picture shows what has actually taken place. We know that the mute is C and C knows that it is C, but in the other prisoner's minds it can be anyone. $\endgroup$ – tyobrien Sep 5 '18 at 15:56

13 Answers 13


It will be


Both A and B can see, what C sees, and that's why they both know that

C knows his hat colour: If C had a white hat, then both A and B would be able to trivially announce their hats. Neither did, and they cannot both be mute, so C must know that his hat isn't white. Because C isn't announcing his colour, both A and B know that C must be the mute.

From there, the problem reverts to the earlier one:

B knows that A isn't the mute (because C is), and also that A isn't seeing three white hats (because A has't announced his hat), so B can decuce that his hat is red.

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    $\begingroup$ Where did you get the fact that C knows his hat color ? $\endgroup$ – Nobody Sep 5 '18 at 6:05
  • $\begingroup$ @casualcoder If C really doesn't know his hat color why should the warden muted C? $\endgroup$ – Viira Sep 5 '18 at 6:55
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    $\begingroup$ @Viira the point of cutting a tongue of a prisoner (others dont know whose mute) alters the strategy of waiting for someone else to speak first as they may be mute and thus cannot speak even if they figured out their hat color $\endgroup$ – Nobody Sep 5 '18 at 7:01
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    $\begingroup$ @SilverCookies if C thinks B is mute, then A's silence reveals C's hat colour. Or if C thinks A is mute, then it's B's silence that gives the colour away. It's an either-or situation. $\endgroup$ – Bass Sep 5 '18 at 9:57
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    $\begingroup$ @SilverCookies, I did the major edit you suggested, and cleaned away my comments that aren't relevant anymore. $\endgroup$ – Bass Sep 5 '18 at 10:38

I'll try another explanation(with same result):

B will tell the right color.

Here are the steps:

C knows, that if his hat was white then A and B would both know, that their color must be RED, as there are 3 white hats in front of them. As none of them speaks, he knows, that his hat is RED.

Next step:

A and B know, that C knows his color. As he does not say it, they know: He is the one without tongue.

So: Who talks?

Now B knows, that if his hat was WHITE, then A would announce his own color to be RED (as he knows, that A is not the mute one). So he now correctly assumes his hat to be RED.

  • $\begingroup$ I do really like this puzzle, i've seen it before but its still a good one, however... if this situation somehow happened in real life, do you not think someone would panic speak out of turn and guess wrong and they'd all end up dead. for the correct answer to this to work requires everyone to understand the riddle already and "play their part" and hope that B is able to figure it out, or is not colour blind? $\endgroup$ – Blade Wraith Sep 5 '18 at 10:22
  • $\begingroup$ @BladeWraith You may find this interesting: puzzling.stackexchange.com/q/58903/40853 $\endgroup$ – ibrahim mahrir Sep 6 '18 at 12:38

C, the mute one, sees there are two white hats in front of them
even if C were not mute, at least one of the people behind them would not be mute. If C's hat were white, both of these people would have red hats, and a nonmute would guess their own hat as red. so, C knows their hat is red after one "tick", and both people behind C know this.

B knows that at least one of A and C are not mute. B can see that C can figure out their own hat colour after one "tick" of silence. and B also knows that, if their hat were white, A would know their own hat colour. When A does not immediately say their own colour, B does not either, nor does C. C now knows their own hat colour, but does not announce it. however, it is debatable whether it is possible for the prisoners to decide when someone really should have said their own hat colour by now. if they can do this, then B should be able to guess their own hat his red

alternately, for lateral thinking:

there is no explicit penalty for guessing wrong. A guesses their hat is red. they do this, because upon them not being immediately freed, this would mean they are wrong, which gives B and C knowledge of a third white hat, allowing both of them to figure out their own hat is red, and guess correctly

  • $\begingroup$ Good job on the extra alternative thinking, but I meant to note there is a penalty for guessing wrong. I added it to my question. $\endgroup$ – tyobrien Sep 5 '18 at 3:15
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    $\begingroup$ @tyobrien while you're changing the question a little bit, perhaps the warden periodically coming into the room to ask for guesses would be helpful for the logic of the first approach $\endgroup$ – Destructible Lemon Sep 5 '18 at 3:18
  • $\begingroup$ I considered it, but for simplicity and to not lead the logic I decided not to. But I might add that each prisoner is infallibly logical and they have unlimited time to figure a solution. $\endgroup$ – tyobrien Sep 5 '18 at 3:24
  • $\begingroup$ Lateral thinking aside, I believe that this should be the correct answer. $\endgroup$ – Xenocacia Sep 5 '18 at 7:48

The person to talk is



After A is silent for a while, B knows that either his hat is red (in which case A does not know his hat colour) or A is mute. B may now reason that for C the situation is clear as well: C sees two white hats, so he knows, that if his hat were white, A and B would instantly know their colour and one of them could speak. So by C's continued silence, he can deduce that C is mute, and A therefore is not, so his hat is red.


my solution:

B waits for a while, then announces he has a red hat. He reasons correctly that if he had a white hat, then A would know that A wore a red hat. When A doesn't say anything, B knows that A's hat is white, therefore his must be red. Note: C could also reason this way, but C is mute.

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    $\begingroup$ Thanks for the response, but B doesn't know that A isn't mute. $\endgroup$ – tyobrien Sep 5 '18 at 2:41

If there was no mute;

$B$ would announce he had a red hat after a while. The only possible condition where $A$ cannot know his hat color when $A$ sees two red and two white hats and since $B$ can see one red and two white hats, he can conclude this.

But there is a mute condition, so

$B$ will not able to deduce easily whether if he had red or white hat since $A$ could be mute too. He needs to wait a bit more. $A$ would know that but since $A$ does not know his hat color he would not able to say anything. $A$ could have a red hat but can be mute then $B$ would have a white hat, etc.

After a while,

B would need to announce that he had a red hat since no one would able to deduce their own hat's color except him. If muting has been done randomly, $B$'s surviving possibility would be $93.3\%$;

with the equation below:

$B$'s survivability chance: $\frac{4}{5}+\frac{1}{5}\cdot \frac{2}{3}=\frac{14}{15}$ without $C$ considered.


$\frac{4}{5}$ is the chance that $A$ is not muted and cannot deduce so $B$ has red hat for sure and $\frac{1}{5}\cdot \frac{2}{3}$ $A$ is muted but still $B$ had red hat.


$B$ notices that $C$ also knows that his hat is red. Because if he had white hat, $A$ and $B$ would know that they had red hat for sure and at least one of them would shout (even one of them was mute), since none of them said anything, $B$ would understand that $C$ is the one who is mute and $A$ is not muted! after a while, $C$ was supposed to say his hat color since nobody said anything and if $C$ had white hat both $B$ and $A$ would know their hat color.


$B$ realizes that $C$ is the silent one! So he has just become sure that he has a red hat now!


Following the answer by SteveV which was

B waits for a while, then announces he has a red hat. He reasons correctly that if he had a white hat, then A would know that A wore a red hat. When A doesn't say anything, B knows that A's hat is white, therefore his must be red. Note: C could also reason this way, but C is mute.

However, B can only infer this result if they know that A is not mute.

B can infer that A is not mute because if A were mute, C would not be mute and given that C knows at least one of A or B are not mute, C could infer that their own hat must be Red because both A and B did not answer. If C's hat were white both A and B would know their hats must be Red and the non-mute one would give an answer. However, since C does not answer, B knows C must be mute and thus A is not mute.

Therefore the solution should be

B can announce that their hat is Red.


I think the right answer has already be given, but since it has the lateral-thinking tag, could it be



"F is completely hidden from view", which mean the guards can't see him, and he's free to take off his hat and check the color. The "buried to their necks" part make it quite hard to take off and put back his hat, but it's still possible!


The answer is



B can see two white hats in front, and can therefore figure out that if his hat was white then A would be speaking up to say that A's own hat was red. Because A remains silent, B knows that his hat must not be white.

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    $\begingroup$ Perhaps A remains silent because he is mute? $\endgroup$ – Mixxiphoid Sep 6 '18 at 11:48
  • $\begingroup$ @Mixxiphoid yes, I thought of that after reading other answers. $\endgroup$ – Chris Peacock Sep 6 '18 at 12:07

There's a lot of answers already but I found some of them hard to follow, so here's mine.

B reasons as follows:
1. If C is not mute, he will reason that he is red as as follows:

a) "There are two white hats in front of me"
b) "If my hat is white, both A and B will see three white hats"
c) "Therefore A and B will both know themselves to be red"
d) "Therefore whichever of A and B is not mute will announce 'red'"
e) "No-one has announced 'red'"
f) "Therefore I am red"

2. C has not announced "red" therefore he is mute.
3. Therefore A is not mute.
4. If my hat were white, A would see three white hats and announce 'white'
5. A has not done this.
6. Therefore I am red.

So B announces his hat colour.


Logic 1

If C, D, E have the same colour hats then A, B both know their hat colour.
Only one of them can be mute so the other can say.
If neither say then everyone knows C, D, E aren't all the same colour.

Logic 2

D, E are the same colour go to Logic 2.1
D, E are different colours go to Logic 2.2

Logic 2.1

If D, E are the same then C knows their colour.
If C isn't mute they can say.
If they don't say anything they must be mute.

Logic 2.2

If D, E are different C will have said nothing, then A, B will have said nothing so D knows he's different to E.
If D isn't mute theu can say.
If D says nothing they must be mute.

Logic 3

A can either see 3 hats the same colour or 2 hats of each colour.
If they can see 3 the same they can say the other colour since C or D is the mute.
If they can see 2 of each they will say nothing.

Logic 4

B now knows A can see 2 of each so they must be different to whichever colour they can see 2 of.
B can say their colour as C or D is the mute.

  • $\begingroup$ No idea why spoilers aren't working $\endgroup$ – Sam Dean Sep 5 '18 at 10:52
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    $\begingroup$ I proposed an edit. You have to prepend >! to every line. $\endgroup$ – EightAndAHalfTails Sep 5 '18 at 10:54



B knows that A is:

1. Mute
2. They sees 2 Red hats and 2 White hats

Conclude that:

"If A is not mute my hat is Red"

B is also aware that C:

1. is mute
2. is not mute, in which case either A or B are not mute(since only one person is). Therefore, if C hat was White, either A or B would correctly guess their hat, so C can deduce that they have a Red hat

This means that C can determine their hat color, however, B notes that C is not talking, therefore they must be mute
Finally B thinks: "If C is mute, A cannot be the mute so my hat is Red"


B has a Red hat

  • $\begingroup$ Can I have some help with the formatting? I cannot hide the explanation $\endgroup$ – SilverCookies Sep 5 '18 at 10:11
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    $\begingroup$ There you go, I proposed an edit. You have to prepend >! to each line and also end each line with two spaces. $\endgroup$ – EightAndAHalfTails Sep 5 '18 at 10:41

EDIT: For certain it could be:



A sees 2 red and 2 white hats, so its hat could be red or white. So it thinks there is a pattern, so announces it's hat color is white (it thinks there can't be 3 red hats together ;) )

I think it is



B waits for A to speak but it's not speaking, so that means either A is mute or not seeing 3 white hats in front of it. So, B is the only one that can make a guess that A is not mute so it will say it's hat color is Red.

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    $\begingroup$ I think the idea is for one of the individuals to reach a logical conclusion about their hat color without guessing $\endgroup$ – SilverCookies Sep 5 '18 at 12:04

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